inverselaplace 1+1/(s(s+2))

解答

的拉普拉斯逆变换

1 + \frac{1}{s ( s + 2 )}

δ (t) + \frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

求解步骤

L^{- 1} {1 + \frac{1}{s ( s + 2 )}}

将

\frac{1}{s ( s + 2 )}

用部份分式展开:

\frac{1}{2 s} - \frac{1}{2 ( s + 2 )}

= L^{- 1} {1 + \frac{1}{2 s} - \frac{1}{2 ( s + 2 )}}

利用逆拉普拉斯变换的线性特性:

对于函数

f (s), g (s)

和常数

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {1} + L^{- 1} {\frac{1}{2 s}} - L^{- 1} {\frac{1}{2 ( s + 2 )}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1}{2 s}} : \frac{1}{2} H (t)

L^{- 1} {\frac{1}{2 ( s + 2 )}} : \frac{1}{2} e^{- 2 t}

= δ (t) + \frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

d er i v a t i v e \frac{1 - 4 x}{2 + x}

\int_{0}^{1} π (2 x)^{2} d x

\int \frac{2 s - 8}{( s ^{2} - 8 s + 25 ) ^{2}} d s

\frac{9 y ^{2} - t ^{2}}{y ^{5}} \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

\frac{d y}{d x} - y = e^{x} \cdot y^{2}